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Is a quotient group a subgroup?

I am trying to solve the question below, and if the answer to above is "yes," then my proof will follow below.

I am trying to prove that a cyclic group G, has a cyclic quotient group G/N. Where N is a normal subgroup.

A SUBGROUP OF A CYCLIC GROUP IS CYCLIC. By https://proofwiki.org/wiki/Subgroup_of_Cyclic_Group_is_Cyclic

So if

(1) G is cyclic

(2) G/N is a subgroup

(3) G/N is a cyclic subgroup

(4) The Quotient group G/N is cyclic

I am unsure about my reasoning in step (2), Is this a valid assumption?

Shaun
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E.J Humphrey
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    No, $G/N$ is not a subset of $G$, so of course it's not a subgroup. The elements of $G/N$ are cosets $gN$, which are not elements of $G$. (Though, if $G$ is cyclic, then $G/N$ is isomorphic to a subgroup of $G$.) – anon Oct 31 '17 at 23:02
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    Sometimes, but not always, there is a subgroup of $G$ isomorphic to $G/N$, but it is not guaranteed. When such a subgroup exists and commutes with the quotient map, the quotient is said to have a "section." – Michael Burr Oct 31 '17 at 23:05
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    Hint: Let $g$ be a generator of $G$. Prove that $gN$ is a cyclic generator of $G/N$. – Michael Burr Oct 31 '17 at 23:06
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    This should be reopened, how can it be off-topic? Just edit title to add "isomorphic to a subgroup". Abelian: https://math.stackexchange.com/questions/64982/is-every-quotient-of-a-finite-abelian-group-g-isomorphic-to-some-subgroup-of Counter example at: https://www.mathcounterexamples.net/is-the-quotient-group-of-a-finite-group-always-isomorphic-to-a-subgroup/ – Ciro Santilli OurBigBook.com Jun 20 '20 at 21:24

2 Answers2

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No, quotients are not subgroups. For example if $G = \mathbb Z$ and $N = 2\mathbb Z$ (the even integers) then $G/N = \mathbb Z/2\mathbb Z$ is cyclic of order $2$, but $\mathbb Z$ does not have any finite subgroups other than the trivial subgroup $0$.

It happens to be true that every quotient of a finite cyclic subgroup is isomorphic to a subgroup of that cyclic group, but you prove that fact by first proving that quotients of cyclic groups are cyclic, thus you cannot use this fact while proving that quotients of cyclic groups are cyclic.

Jim
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Let $G$ be a cyclic group. Let $N$ be a subgroup of $G$. Note that all cyclic groups are Abelian, so all subgroups are normal (think about it). Then $G = \langle g\rangle$ for some $g\in G$. It sounds like you want to show that $G/N$ is also cyclic. As a hint see if you can show that $G/N = \langle gN\rangle$.

Note that the quotient $G/N$ is a set of cosets. As such it is not a subgroup of $G$. It isn't even a subset. A subgroup is in particular a subset.

Thomas
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